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*From*: Bowman_David/tiger_mpc@tiger.gtc.georgetown.ky.us*Date*: Tue, 2 Apr 96 17:46:36 -0500

Vince S. writes:

There seems to be at least one more way to get the correct answer:

Then there are two ways to get the right answer:

Method A: simple physics in the earth frame

Method B: the Lorentz transformations.

Method C: Treat the problem as a case of the relativistic Doppler shift.

In this method consider the "light" signal from the tail of the spacecraft

as a (outgoing spherical) radio wave whose wavelength is 200 m (the rocket's

length) in the rocket's rest frame. Calculate the period of this wave; it is

the time in the answer to part a of the problem. Now consider the period of

this wave as viewed from an Earthbound observer watching the rocket recede

from her. This period is lengthened (red-shifted) by a Doppler factor of

sqrt((1+0.95)/(1-0.95)) = sqrt(39) = 6.245. (Note that

4.166X10^-6 s = 6.245 x 6.671x10^-7 s). The red-shift formula rather than

the blue-shift formula is used because for this problem we assume that the

earthbound observer is located at the rocket tail (source) when the initial

wave-crest is emitted and that source is moving away from the earthbound

observer as the wave cycle is being emitted. The tail is to be coincident

with the earthbound observer at the emission of the initial wave-crest since

the time interval measured in both frames should be measured with coincidently

reset clocks.

David Bowman

Georgetown College

dbowman@gtc.georgetown.ky.us

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